• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Determination of the formula of Hydrated Iron(II) Sulphate Crystals

Extracts from this document...


Determination of the formula of Hydrated Iron(II) Sulphate Crystals Hydrated FeSO4 can come in many hydrated forms. Using two methods for better accuracy and to make a comparison, I will deduce the number of water molecules per mole of a hydrated iron(II) sulphate sample. Method 1: FeSO4.nH2O(s) � FeSO4(s) + nH2O(l) =dehydrating the FeSO4 by evaporating the water and calculating the mass lost. Hazards in this experiment include the flame from the Bunsen and the possibility of the crucible shattering. By using a gentle heat for two minute intervals and the lid slightly ajar (but not so the contents might spit) the crucible shouldn't spit and the water can escape. The reactant (FeSO4) is not dangerous, unless consumed in large doses. Results: Crucible, lid and FeSO4= 21.76g FeSO4= 1.41g Heating Mass after heating(g) Overall loss(g) 1 21.10 0.66 2 21.10 0.66 3 21.10 0.66 4 21.10 0.66 All four results show one and no further mass change meaning the water has been evaporated leaving just the iron(II) sulphate. Analysis: Out of the 1.41g sample 0.66g was water and the other 0.75g being FeSO4. By working out the molar ratio the relative molar mass are taken into account and the amount of water present in the hydrated FeSO4 can be determined. ...read more.


ions, in the hydrated iron(II) sulphate, into iron(III) ions: 1. Fe2+(aq) � Fe3+(aq) + e-, and 2. MnO4-(aq) + 8H+(aq) +5e- � Mn2+(aq) + 4H2O(l). To put these half equations into a full ionic equation the electrons must cancel out on either side. To do this the oxidisation equation (1) is multiplied by 5 and combined with the reduction equation (2) to give the complete redox ionic equation: MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) � Mn2+(aq) + 5Fe3+(aq) + 4H2O(l). From this equation the molar ratio between potassium permanganate and iron(II) sulphate is shown to be 1:5, so (2.2275�10-4)�(5) moles of hydrated iron(II) sulphate were present in 25cm3, and (2.2275�10-3) � (5) were present overall in the whole 250cm3 solution. Moles of iron(II) sulphate = (2.2275�10-3) � (5) = 1.11375�10-2 moles. The mass of the iron(II) sulphate that reacted was moles � molecular mass; (1.11375�10-2) � (151.9) = 1.69g.(3.s.f) As the water in the hydrated iron(II) sulphate weighed at the start played no part in the reaction, subtracting the mass of the anhydrous iron(II) sulphate from the initial hydrated form will give the mass of water present in the sample; 3.05g - 1.69g = 1.36g of water. ...read more.


This method generally relies on chemistry instead of maths and the only variable that could have caused the off answer was again the scales. Carrying out this experiment on a larger scale would be far too dangerous in a school due to the use of fire. The problem with method one being more in touch with chemistry is the risk of procedural errors. On heating the melanterite the substance changed in colour from blue to white indicating the water had been eliminated. However there were also areas of dark red which indicates a further reaction. The heat was too strong and iron(III) oxide was formed with sulphur di- and trioxide: 2FeSO4(s) � Fe2O3(s) + SO2(g) + SO3(g). These gases also escaped with the water and gave the impression that there was more water present as some sulphur and oxygen escaped as well. Heating the melanterite more gently and patiently for less time would make sure this didn't happen. If this didn't happen I am confident that this method would be more accurate than a titration as it is more purely chemistry instead of maths, which holds a greater chance of carrying errors over, despite my results suggesting otherwise. ?? ?? ?? ?? Rob H ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE Aqueous Chemistry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Aqueous Chemistry essays

  1. Peer reviewed

    Formula of a hydrated salt

    4 star(s)

    Concentration = 0.01 mol dm-3 Volume = 17.50 cm3 / 1000 = 0.0175 dm3 Therefore 0.01 mol dm-3 x 0.0175 dm3 = 1.75 x 10-4 mol Now using the equation: 5Fe2+ + MnO4- + 8H+ = 5Fe3+ + Mn2+ + 4H2O, I know that there are 5 times the amount of Fe2+ (FeSO4)

  2. The aim of this practical experiment is to calculate the formula of hydrated Iron(II) ...

    Calculating the number of moles in the solution and the can allow as to calculate the formula. Hypothesis Iron (II) Sulphate salt normally exists as a hydrated salt. Hydrated salt contains a fixed amount of water and in most case, the number of moles of water is discrete.

  1. How much Iron (II) in 100 grams of Spinach Oleracea?

    Place the conical flask on a white tile. 7) While swirling the conical flask, titrate the Potassium Manganate (VII) (aq) into the conical flask containing Iron (II) Ammonium Sulphate (aq) and Sulphuric Acid (aq) until the first trace of a pink colour appears.

  2. Determining the purity of Iron Wool.

    A titration method relevant to an acid-base reaction will be used to determine the quantity of hydrochloric acid needed to bring 25cm3 sodium carbonate solution to equilibrium. For two solutions to reach equilibrium means they are balanced in proportions to one other.

  1. Synthesis of Iron (II) Sulphate FeSO4

    Total weight of crystals of Iron (II) Sulphate (FeSO4) is 4.3 gram The percentage yield of the salt: The number of mol of Sulphuric acid (H2SO4) used: 40ml x 1mol/dm3 = 0.040dm3 x 1mol/dm3 = 0.040 mol The reacting equation can be represented as Fe + H2SO4 --> FeSO4 + H2 - From the equation, 1 mol of Sulphuric acid reacts totally with excess Iron fillings (Fe)

  2. A2 Chemistry -Assessed Practical In theory the remaining mass after the heating will be ...

    you can see that there are 5 moles of Fe2+ taking part in the reaction so: 5 x 2.097x10-� = 1.103125 mol This method has only calculated this for 25cm3 of solution but we need 250 cm3 so we multiply by 10.

  1. Determine the relative formula mass and the molecular formula of succinic acid

    22.40 22.45 22.42 22.42 Initial Volume ( cm3) 0.00 0.00 0.00 0.00 Volume of titre ( cm3) 22.40 22.45 22.42 22.42 Discussion: The gram concentration of the acid solution =1.65/250cm3 =6.6g.dm-3 Assume that n=3, Mr of succinic acid will be 132.

  2. The Estimation of Iron (II) and Iron (III) in a Mixture Containing Both.

    o The titration must be set up now with the burette attached to the boss and clamp with a conical flask above a white tile below. o The pipette must be cleaned out using a sample of the iron solution.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work